Question

Which of the following are paramagnetic in nature?

• A. ${\mathrm{O}}_2$
• B. ${{\mathrm{O}}_2}^{+}$
• C. ${{\mathrm{O}}_2}^{-}$
• D. All of these

Answer 1

The correct option is D

All of these

Explanation for correct option

(D) All of these

Magnetic Nature

• Materials having unpaired electrons are paramagnetic and attracted to a magnetic field.
• Examples of paramagnetic: aluminum, gold, and copper.
• Materials containing paired electrons are diamagnetic and repel a magnetic field.
• Examples of diamagnetic: copper, zinc, silver.

${\mathbf{O}}_{\mathbf{2}}$

• An atom having one or more unpaired electrons is said to be in a paramagnetic.
• According to the molecular orbital theory, the electronic configuration of Oxygen molecule$\left({\mathrm{O}}_2\right)$ is $\left(\sigma 1s{)}^2{\left({\sigma }^{\star }1s\right)}^2\right(\sigma 2s{)}^2{\left({\sigma }^{\star }2s\right)}^2(\sigma 2p_z{)}^2\left(\pi 2p_x^2=\pi 2p_y^2\right)\left({\pi }^{\mathit{\star }}2p_x^1={\pi }^{\mathit{\star }}2p_y^1\right)$.
• As it has two unpaired electrons, it is paramagnetic in nature.

${{\mathbf{O}}_{\mathbf{2}}}^{\mathbf{-}}$

• ${{\mathrm{O}}_2}^{-}$ has a bond order of $1.5$ and three electrons in the ${\pi }^{*}$orbitals.
• According to the molecular orbital theory, the electronic configuration of ${{\mathbf{O}}_{\mathbf{2}}}^{\mathbf{-}}$ is:$\left(\sigma 1s{)}^2{\left({\sigma }^{\star }1s\right)}^2\right(\sigma 2s{)}^2{\left({\sigma }^{\star }2s\right)}^2(\sigma 2p_z{)}^2\left(\pi 2p_x^2=\pi 2p_y^2\right)\left({\pi }^{\mathit{\star }}2p_x^2={\pi }^{\mathit{\star }}2p_y^1\right)$.
• The average bond order of the Lewis structures is $1.5$, and they have an unpaired electron.
• As it has one unpaired electron, it is paramagnetic in nature.

${{\mathbf{O}}_{\mathbf{2}}}^{\mathbf{+}}$

• According to the molecular orbital theory, the electronic configuration of ${{\mathbf{O}}_{\mathbf{2}}}^{\mathbf{+}}$ is:

$\left(\sigma 1s{)}^2{\left({\sigma }^{\star }1s\right)}^2\right(\sigma 2s{)}^2{\left({\sigma }^{\star }2s\right)}^2(\sigma 2p_z{)}^2\left(\pi 2p_x^2=\pi 2p_y^2\right)\left({\pi }^{\mathit{\star }}2p_x^1\right)$.

• Since this orbital has the highest energy, the electron would be taken from it.
• ${{\mathrm{O}}_2}^{+}$ is paramagnetic because it has an unpaired electron.

.therefore, the correct option is (D) All of these.